# Circuit wanted to run 12VDC LEDs from Low Voltage landscape circuit

I have some 1w 12VDC Weatherproof LEDs that I'd like to use for lighting my pond, where I have an existing 12VAC lighting circuit. I understand that the voltage from the landscape light circuit could run as high as 14V, although I haven't measured it yet. What can I use (that I could reasonably make weatherproof) that would allow me to run 3 (perhaps 4) of these from this supply?

• I feel like we had a question almost identical to this one that got migrated to home improvement. I am having a hard time finding it now though. Jul 7, 2011 at 21:02

In order of increasing preference:

• 1 A series diode to a capacitor

• 2 A bridge rectifier to a capacitor.

• 3 A bridge rectifier to capacitor, with a series resistor to LEDs (best, a series resistor in each string).

• 4 Overkill: Bridge rectifier, capacitor, series regulator or resistor + shunt zener

LED strings in parallel in each case.

Because: The "12 Volt" weatherproof LEDs are almost certainly between 1 and 3 LEDs in series plus a resistor. This combination allows some tolerance as increasing applied voltage causes increasing current which leads to more drop across the series resistor, so the LEDs voltage drop increases only "modestly" with increasing voltage.

Option 3 is liable to be the most effective. If the LDs are standard single 5mm or 3mm LEDS the current drain is probably designed at 20 mA. If they are series parallel strings then they could have almost any rating and you need either to look at a data sheet, or measure current at 12V or use trial and error.

The 12 VAC supply, when full wave rectified with a "bridge" rectifier, will produce about 16 VDC. Actual voltage depends on length of wiring, load from other lights, mains voltage fluctuations etc BUT 16 VDC is a good first guess. So you need to drop about 16-12 = 4 volts. Series resistor required is $R = \frac{V}{I}$. At 20 mA this would be $R = \frac{4}{0.02} \approx 200\Omega$. Nearest standard "E12" values are 180R (180 Ohms) or 220R (220 Ohms). Chances are that 180R would be fine and even 270R or 330R or 150R or 120R would be good enough.

The capacitor needs to be large enough to reduce ripple from raw rectified AC and as small as possible for \$ and size reasons. The whole arrangement can be placed in a minibox or potted in epoxy or ... . For outdoor protection I sometimes use a pair of suitably sized soft-drink bottles cut such that one pushes over the other and pressed together and located so water has a "hard job" entering. This can be as capable and aesthetic an arrangement as your skill and caring allow.

The bridge rectifier can be almost any such - small ones rated at 100's of mA to 1A to much more are available from electronic components suppliers (Frys?, Radio Shack?, Digikey, Farnell, Mouser, ...).

Power dissipation is mainly in the resistor and is $I^2 * R$ or $\frac{V^2}{R}$ or $V * I$.
(V = voltage drop across R, I = LED current.)
e.g. for 180R and 20 mA Power:

$R = I^2 * R = 0.02^2 * 180 = 0.072 \;\;$ Watts = trivial.

If your 12V LED is e.g. 3 x 1 Watt rated at 350 mA then power in series:

$R = V * I = 4 * 0.35 = 1.4 \;\;$ Watts.

This is less trivial. Potting that in epoxy without care may led to "problems". A suitably power rated resistor is required in each case.

Capacitor is sized to allow acceptable ripple during about one half mains cycle. Roughly:

Voltage Drop $= V = t * \frac{I}{C} \;\;$ Volts
(t = seconds, V = drop in time t, I = current drawn, C = capacitance in farads.

Rearranging $C= t * \frac{I}{V}$ For 60 Hz t ~= 8 mS, i = 20 mA say, V = 1 Volt drop say. C = 0.00016 F = 160 uF. i.e. you need ABOUT 160 uF per 20 mA LED. You MAY be happy with no cap at all and full wave DC on LEDS. You MAY see too much flicker with e.g. 100 uF per 20 mA LED. Chances are that a 1000 uF, 16 Volt capacitor will be fine for 3 x 20 MA LEDS and that smaller will be OK.

Sensibility check After building, measure Vdc across LEDs . Should be ~~~~ 12 Volts.

All the above said, solution 1. or 2. may work well enough for you.

Added: The extra information and page you referred to suggests that they may use a SR (switching regulator) (probably a buck converter) internally. The ones shown state that they are rated for 8 to 24 V, which is a very wide range for a linear regulator due to efficiency considerations and far beyond what you can sensibly achieve with a resistor only. The current rating of 156.3 mA at 12V to give 71 lumen output also strongly suggests a SR. 12v x 156 mA = 1.9 Watt or 38 lumen/Watt. This is modest but realistic in real world applications.

So

• Contact the suppliers and ask them and/or

• Try the 2nd circuit with no cap. Flicker may be objectionable.

• Add a capacitor as in 2 & 3. At 156 mA, using my 160 uF per 20 mA rule of thumb, for 1 Volt drop you'd want

156 mA / 20 mA x 160 uF = 1248 uF, so a 16V, 1000uF aluminium electrolytic would probably be fine.

• This is a single "high power" LED. Current is listed here as 155mA; I also see that volage range is listed as 8-24, which I hadn't noted before. I'd like to run 3 of them in parallel, so total load would be about 465mA. (Mine are older, and are not labelled AC/DC as the one in the photo is)
– TomG
Jul 7, 2011 at 3:20
• In the last section, you say $C= 0.008 * \frac{0.02}{1}$. I get 0.00016 for that, not 0.0016. Is that a typo? You then go on to say 1600, i.e about 160. I think there is a transposed zero in there. Jul 7, 2011 at 7:31
• Yes 1600 uF was an error (factor of 10 too high) which I corrected to 160 uF and missed 2 instances. Jul 7, 2011 at 9:03
• @Tom - That link shows a single high-power LED with a turn-on voltage of 8V. No single LED has that high a forward voltage, and no set of series LEDs will tolerate a 4V rise (8V to 12V, much less 8V to 24V) if there's no resistor. It's a single package containing a resistor and probably 3 LEDs in series. Jul 7, 2011 at 16:55
• @Russel - Off by a factor of 10 is an engineer's version of an off by one error. Jul 7, 2011 at 16:56